Outline

The proof uses the convenient fact that M⁢Aκ holds as long as it holds for all partial orders smaller than κ. Given the conditions on κ, there are at most κ names for these partial orders. At each step in the forcing, we force with one of these names. The result is that the actual generic subset we add intersects every dense subset of every partial order.

Construction of Pκ

Q^α will be constructed by induction with three conditions: |Pα|≤κ for all α≤κ, ⊩PαQ^α⊆ℳ, and Pαsatisfies the ccc. Note that a partial ordering on a cardinalλ<κ is a function from λ×λ to {0,1}, so there are at most 2λ<κ of them. Since a canonical name for a partial ordering of a cardinal is just a function from Pα to that cardinal, there are at most κ2λ≤κ of them.

At each of the κ steps, we want to deal with one of these possible partial orderings, so we need to partition the κ steps in to κ steps for each of the κ cardinals less than κ. In addition, we need to include every Pα name for any level. Therefore, we partion κ into ⟨Sγ,δ⟩γ,δ<κ for each cardinal δ, with each Sγ,δ having cardinalityκ and the added condition that η∈Sγ,δ implies η≥γ. Then each Pγ name for a partial ordering of δ is assigned some index η∈Sγ,δ, and that partial order will be dealt with at stage Qη.

Formally, given Q^β for β<α, Pα can be constructed and the Pα names for partial orderings of each cardinal δ enumerated by the elements ofSα,δ. α∈Sγ,δ for some γα and δα, and α≥γα so some canonical Pγα name for a partial order ≤^α of δα has already been assigned to α.

Since ≤^α is a Pγα name, it is also a Pα name, so Q^α can be defined as ⟨δα,≤^α⟩ if ⊩Pα⟨δα,≤^α⟩ satisfies the ccc and by the trivial partial order ⟨1,{⟨1,1⟩}⟩ otherwise. Obviously this satisfies the ccc, and so Pα+1 does as well. Since Q^α is either trivial or a cardinal together with a canonical name, ⊩PαQ^α⊆ℳ. Finally, |Pα+q|≤∑n|α|n⋅(supi⁡|Q^i|)n≤κ.

Proof that M⁢Aλ holds for λ<κ

Lemma: It suffices to show that M⁢Aλ holds for partial order with size ≤λ

Proof.

Suppose P is a partial order with |P|>κ and let ⟨Dα⟩α<λ be dense subsets of P. Define functions fi:P→Dα for α⁢κ with fα⁢(p)≥p (obviously such elements exist since Dα is dense). Let g:P×P→P be a function such that g⁢(p,q)≥p,q whenever p and q are compatible. Then pick some element q∈P and let Q be the closure of {q} under fα and g with the same ordering as P (restricted to Q).

Since there are only κ functions being used, it must be that |Q|≤κ. If p∈Q then fα⁢(p)≥p and clearly fα⁢(p)∈Q∩Dα, so each Dα∩Q is dense in Q. In addition, Q is ccc: if A is an antichain in Q and p1,p2∈A then p1,p2 are incompatible in Q. But if they were compatible in P then g⁢(p1,p2)≥p1,p2 would be an element of Q, so they must be incompatible in P. Therefore A is an antichain in P, and therefore must have countable cardinality, since P satisfies the ccc.

By assumption, there is a directed G⊆Q such that G∩(Dα∩Q)≠∅ for each α<κ, and therefore M⁢Aλ holds in full.
∎

Now we must prove that, if G is a generic subset of Pκ, R some partial order with |R|≤λ and ⟨Dα⟩α<λ are dense subsets of R then there is some directed subset of R intersecting each Dα.

If |R|<λ then λ additional elements can be added greater than any other element of R to make |R|=λ, and then since there is an order isomorphism into some partial order of λ, assume R is a partial ordering of λ. Then let D={⟨α,β⟩∣α∈Dβ}.

Take canonical names so that R=R^⁢[G], D=D^⁢[G] and Di=D^i⁢[G] for each i<λ and:

⊩PκR^⁢ is a partial ordering satisfying ccc andD^⊆λ×λ⁢ and Dα^⁢ is dense in ⁢R^

For any α,β there is a maximal antichain Dα,β⊆Pκ such that if p∈Dα,β then either p⊩Pκα≤R^β or p⊩Pκα≰R^β and another maximal antichain Eα,β⊆Pκ such that if p∈Eα,β then either p⊩Pκ⟨α,β⟩∈D^ or p⊩Pκ⟨α,β⟩∉D^. These antichains determine the value of those two formulas.

Then, since κcf⁡κ>κ and κμ=κ for μ<κ, it must be that cf⁡κ=κ, so κ is regular. Then γ=sup({α+1∣α∈dom(p),p∈⋃α,β<λDα,β∪Eα,β)<κ, so Dα,β,Eα,β⊆Pγ, and therefore the Pκ names R^ and D^ are also Pγ names.

Lemma: For any γ, Gγ={p↾γ∣p∈G} is a generic subset of Pγ

Proof.

First, it is directed, since if p1↾γ,p2↾γ∈Gγ then there is some p∈G such that p≤p1,p2, and therefore p↾γ∈Gγ and p≤p1↾γ,p2↾γ.

Also, it is generic. If D is a dense subset of Pγ then Dκ={p∈Pκ∣p≤q∈D} is dense in Pκ, since if p∈Pκ then there is some d≤p↾, but then d is compatible with p, so d∪p∈Dκ. Therefore there is some p∈Dκ∩Gκ, and so p↾∈D∩Gγ.
∎

Since R^ and D^ are Pγ names, R^⁢[G]=R^⁢[Gγ]=R and D^⁢[G]=D^⁢[Gγ]=D, so

V⁢[Gγ]⊨R^⁢ is a partial ordering of ⁢λ⁢ satisfying the ccc andDα^⁢ is dense in ⁢R^

Then there must be some p∈Gγ such that

p⊩PγR^⁢ is a partial ordering of ⁢λ⁢ satisfying the ccc

Let Ap be a maximal antichain of Pγ such that p∈Ap, and define ≤^* as a Pγ name with (p,m)∈≤^* for each m∈R^ and (a,n)∈≤^* if n=(α,β) where α<β<λ and p≠a∈Ap. That is, ≤^*⁢[G]=R when p∈G and ≤^*[G]=∈↾λ otherwise. Then this is the name for a partial ordering of λ, and therefore there is some η∈Sγ,λ such that ≤^*=≤^η, and η≥γ. Since p∈Gγ⊆Gη, Q^η⁢[Gη]=≤^η⁢[Gη]=R.

Since Pη+1=Pη*Qη, we know that GQη⊆Qη is generic since http://planetmath.org/node/3258forcing with the composition is equivalent to successive forcing. Since Di∈V⁢[Gγ]⊆V⁢[Gη] and is dense, it follows that Di∩GQη≠∅ and since GQη is a subset of R in Pκ, M⁢Aλ holds.

Proof that 2ℵ0=κ

The relationship between Martin’s axiom and the continuum hypothesis tells us that 2ℵ0≥κ. Since 2ℵ0 was less than κ in V, and since |Pκ|=κ adds at most κ elements, it must be that 2ℵ0=κ.